Frictional force with appiled force at an angle

[Soniteflash]

Homework Statement

An applied force of 30N acts on a 8kg box at an angle of 40 degrees above the horizontal. What is the acceleration of the block if

=0.1 and μk = 0.05? What is the acceleration of the block without any friction?

Homework Equations

Equation 1
F k = μ k F N Equation 2
F=ma Equation 3

The Attempt at a Solution

Question 2[/B]
I attempted the second question first since it seemed to me easier. First I set up F=ma and plugged in cos(40)*30 since I need the x component of the applied force and then the mass which is 8 kg. Solving it gave me 2.9m/s² which was the correct answer for the question.

Question1
For the first question I began with identifying all the forces acting in the Y-direction since the normal force would be needed to calculate the static friction force with equation 1 : F_n
, F_mg and the y component of the applied force F_a.
Both F_mg and F_ay act in the negative y direction so adding them up would equal F_n. Using trig functions I calculated F_ay to be sin(40)30=19.28N and for
F_mg I got 78.48N. F_n=78.48N + 19.28N= 97.76N

Next I identified all forces acting in the x-direction which is the x-component of the applied force and the static frictional force acting in the opposite direction: F_ax and F_fs.
Then I used the sum of forces in the x- direction and got
F_xsum=F_ax + (-F_fs)
= cos(40)*30N - (0.1*97.76N)
= 22.98N - 9.776N
= 13.204N
So i think that's the remaining force after static friction has been overcome?
Now I am a bit confused on how to continue. Do I now calculate the kinetic friction and the subtract it from 13.204N and then use equation 3 (F=ma) to solve for the acceleration?

The given solutions were: acceleration without friction: 2.9m/s²
acceleration with friction : 2.3m/s²

This is my first post in the homework(first post ever in this forum) section and I apologize if I did not meet all of the Guidelines in this section (self-made graphic is too big?). I hope I didn't put too much stuff into this thread but I thought that if I put less, it would be harder for the forum members to understand what I am struggling with.

 

[Doc Al]

Next I identified all forces acting in the x-direction which is the x-component of the applied force and the static frictional force acting in the opposite direction: Fax and Ffs.
Then I used the sum of forces in the x- direction and got
Fxsum=Fax + (-Ffs)
= cos(40)*30N - (0.1*97.76N)
= 22.98N - 9.776N
= 13.204N
So i think that's the remaining force after static friction has been overcome?

That's not quite how it works. For static friction, you first calculate the max value of static friction (which is what you did). If your applied force in the horizontal direction is greater than that max value, then static friction no longer applies. You have to redo your calculation with kinetic friction.

(If your applied force is less than the max value of static friction, then nothing moves. In that case, the actual static friction will equal the amount of your applied force in the horizontal direction.)

 

[Soniteflash]

Oh, ok yeah now I remember my Physics teacher told me about the maximum static friction force. I totally forgot that static friction either is equal to the applied force or that it is ignored if the applied force exceeds it. OK, so I just calculate the kinetic force and subtract the result from the applied force and then use F=ma to get the acceleration?
Attempt:

fk=μkN = 0.05(97.76)=4.888N
F_ax= 22.98N
Plug and chuck.
F⃗ net_x=ΣF⃗ _x=ma⃗ _x
=F_n - F_k=8kg * a
= 22.98N -4.888N = 8kg * a
=18.092=8a
a= 2.36m/s²

The solution matches the given solution.

 

[Doc Al]

Oh, ok yeah now I remember my Physics teacher told me about the maximum static friction force. OK, so I just calculate the kinetic force and subtract the result from the applied force and then use F=ma to get the acceleration?

You got it. Good job!

 

[Soniteflash]

Yuuuuuss. Thank you sir!